Q. ~Magic~Tentukan turunan pertama dari:[tex] \tt f(x) = \frac{ {9x}^{2} \sqrt{8x} - 3 }{x \sqrt{x} .... Question from @misrokhah439

August 2022 1 38 Report

Q. ~Magic~

Tentukan turunan pertama dari:
[tex] \tt f(x) = \frac{ {9x}^{2} \sqrt{8x} - 3 }{x \sqrt{x} } [/tex]

no copas
no ngasal
menggunakan cara

henriyulianto

Turunan pertama dari
$f(x)=\dfrac{{9x}^{2}\sqrt{8x}-3}{x \sqrt{x} }$
adalah:
$\large\text{$\begin{aligned}f'(x)&=\boxed{\ \bf18\sqrt{2}+\frac{9}{2x^2\sqrt{x}}\ }\end{aligned}$}$
 

Pembahasan

Turunan

Diketahui

$\large\text{$\begin{aligned}&f(x)=\frac{{9x}^{2}\sqrt{8x}-3}{x \sqrt{x} }\end{aligned}$}$

Ditanyakan

Turunan pertama dari $f(x)$

PENYELESAIAN

Cara Pertama: Aturan Pembagian pada Turunan

$\begin{aligned}f(x)&=\left(\frac{g}{h}\right)(x)\\\Rightarrow f'(x)&=\frac{f'(x)g(x)-g'(x)f(x)}{\left[g(x)\right]^2}\end{aligned}$
.....................................

$f(x)=\dfrac{{9x}^{2}\sqrt{8x}-3}{x\sqrt{x}}$

Maka, turunan pertamanya adalah:

$\begin{aligned}f'(x)&=\frac{\left({9x}^{2}\sqrt{8x}-3\right)'x\sqrt{x}-\left(x\sqrt{x}\right)'\left({9x}^{2}\sqrt{8x}-3\right)}{\left(x\sqrt{x}\right)^2}\\&=\frac{\left(9\sqrt{8}\cdot x^{5/2}-3\right)'x^{3/2}-\left(x^{3/2}\right)'\left(9\sqrt{8}\cdot x^{5/2}-3\right)}{x^3}\\&=\frac{\left(9\cdot2\sqrt{2}\cdot\dfrac{5x^{3/2}}{2}\right)x^{3/2}-\dfrac{3x^{1/2}}{2}\left(9\cdot2\sqrt{2}\cdot x^{5/2}-3\right)}{x^3}\\&=\frac{45\sqrt{2}x^3-27\sqrt{2}x^3+\dfrac{9x^{1/2}}{2}}{x^3}\end{aligned}$
$\begin{aligned}f'(x)&=\frac{18\sqrt{2}x^3+\dfrac{9x^{1/2}}{2}}{x^3}\\&=18\sqrt{2}+\frac{9}{2}x^{(1/2\:-\:3)}\\&=18\sqrt{2}+\frac{9}{2}x^{-5/2}\\&=18\sqrt{2}+\frac{9}{{2x}^{5/2}}\\f'(x)&=\boxed{\ \bf18\sqrt{2}+\frac{9}{2x^2\sqrt{x}}\ }\end{aligned}$
$\blacksquare$

Cara Kedua: Aturan Penjumlahan/Pengurangan pada Turunan

$\begin{aligned}f(x)&=(g\pm h)(x)\\\Rightarrow f'(x)&=g'(x)\pm h'(x)\end{aligned}$
.....................................

$\begin{aligned}f(x)&=\frac{{9x}^{2}\sqrt{8x}-3}{x\sqrt{x}}\\&=\frac{9\sqrt{8}\cdot x\cdot x\sqrt{x}-3}{x\sqrt{x}}\\&=9\sqrt{8}x-\frac{3}{x\sqrt{x}}\\f(x)&=18\sqrt{2}x-3x^{-3/2}\end{aligned}$

Maka, turunan pertamanya adalah:

$\begin{aligned}f'(x)&=\left(18\sqrt{2}x-3x^{-3/2}\right)'\\&=\left(18\sqrt{2}x\right)'-\left(3x^{-3/2}\right)'\\&=18\sqrt{2}-\left(-\frac{3}{2}\right)3x^{-5/2}\\&=18\sqrt{2}+\frac{9}{2}x^{-5/2}\\&=18\sqrt{2}+\frac{9}{{2x}^{5/2}}\\f'(x)&=\boxed{\ \bf18\sqrt{2}+\frac{9}{2x^2\sqrt{x}}\ }\end{aligned}$
$\blacksquare$
 